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Current Question (ID: 20350)

Question:

$\text{Find the magnetic field at point } P \text{ due to a straight line segment } AB$ $\text{of length } 6 \text{ cm carrying a current of } 5 \text{ A. (See figure)}$ $\left( \mu_0 = 4\pi \times 10^{-7} \text{ N}\cdot\text{A}^{-2} \right)$

Options:

1. $1.5 \times 10^{-5} \text{ T}$
2. $3.0 \times 10^{-5} \text{ T}$
3. $2.0 \times 10^{-5} \text{ T}$
4. $2.5 \times 10^{-5} \text{ T}$

Solution:

$\text{Hint: } B = \frac{\mu_0 i}{4\pi d} \left( \sin \theta_1 + \sin \theta_2 \right)$ $B = \frac{5}{4 \times 10^{-2}} \left( \frac{3}{5} + \frac{3}{5} \right) \times 10^{-7}$ $= \frac{5}{4} \times 2 \frac{3}{5} \times \frac{10^{-7}}{10^{-2}}$ $B = 1.5 \times 10^{-5} \text{ T}$

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Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}